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(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Given $V = b - ar^2$,we have $E = -\frac{d}{dr}(b - ar

Vedclass · Accepted Answer

$-6 a\varepsilon_0$

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(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Given $V = b - ar^2$,we have $E = -\frac{d}{dr}(b - ar^2) = 2ar$.According to Gauss's Law in differential form,the charge density $ho$ is given by $ho = \varepsilon_0 (
abla \cdot E)$.In spherical coordinates,for a radial field $E(r)$,the divergence is $
abla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 E)$.Substituting $E = 2ar$,we get $
abla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot 2ar) = \frac{1}{r^2} \frac{d}{dr}(2ar^3) = \frac{1}{r^2} (6ar^2) = 6a$.Therefore,the charge density is $ho = \varepsilon_0 (6a) = 6a\varepsilon_0$.

The electrostatic potential inside a charged spherical ball is given by $V = b - ar^2$,where $r$ is the distance from the centre; $a$ and $b$ are constants. Then,the charge density inside the ball is:

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